Captain Kevin has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Brandon and his merciless band of thieves. The Captain has probability $\dfrac{4}{9}$ of hitting the pirate ship, if his ship hasn't already been hit. If it has been hit, he will always miss. The pirate has probability $\dfrac{1}{3}$ of hitting the Captain's ship, if his ship hasn't already been hit. If it has been hit, he will always miss as well. If the Captain shoots first, what is the probability that both the Captain and the pirate hit each other's ships?
Explanation: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is the Captain hitting the pirate ship and event B is the pirate hitting the Captain's ship. The Captain fires first, so his ship can't be sunk before he fires his cannons. So, the probability of the Captain hitting the pirate ship is $\dfrac{4}{9}$ If the Captain hit the pirate ship, the pirate has no chance of firing back. So, the probability of the pirate hitting the Captain's ship given the Captain hitting the pirate ship is $0$ The probability that both the Captain and the pirate hit each other's ships is then the probability of the Captain hitting the pirate ship times the probability of the pirate hitting the Captain's ship given the Captain hitting the pirate ship. This is $\dfrac{4}{9} \cdot 0 = 0$